Question

Find the equation to the hyperbola, whose eccentricity is $\frac{5}{4}$, whose focus is $(a,0)$, and whose directrix is $4x-3y=a$.
Find also the coordinates of the centre and the equation to the other directrix.

Answer 1

let $P(x,y)$ be any point on the hyperbola

and $l_1$ is perpendicular from $P$ on the directrix $l_1$.

Then by definition

$F_{1}P=\frac{c}{q}=eP{l}_1$

$\Rightarrow {F_{1}P}^2={\left(eP{l}_1\right)}^2$

${(x-a)}^2+{(y-0)}^2=\frac{25}{16}\times {\left(\frac{4x-3y-a}{\sqrt{16+9}}\right)}^2$

$16x^2+16y^2+16a^2-32ax=16x^2+9y^2-24xy+a^2-8ax+6ay$

The equation of hyperbola is

${7y}^2{+15a}^2+24xy-24ax-6ay=0$

Distance of $l_1$ from $F_1=\frac{4\left(a\right)-3\left(0\right)-a}{\sqrt{4^2+3^2}}=\frac{3a}{5}=\frac{c^2-q^2}{c}$

$\frac{c}{q}=e$

$\Rightarrow c=\frac{3a{e}^2}{(e^2-1)5}$

$\Rightarrow c=\frac{5a}{3}$

$F_1{F}_2=2c$ and $F_{1}F2$ is perpendicular to $l_1$

$\therefore$ $F_1{F}_2$ is at a distance of $2c$ at slope $=\frac{-3}{4}$ towards $l_1$

on solving we get $F_2=(a-2\times \frac{4c}{5},0-2\times \frac{3c}{5})=(\frac{-5a}{3},2a)$

Centre of the hyperbola is the midpoint of $F_1{F}_2$ ie. at $(\frac{-a}{3},a)$

Let the equation of directrix $l_2$ be $4x-3y=k\because l_{1}and{l}_2$ are parallel with distance between them equal to $\frac{2q^2}{c}=\frac{32c}{25}$

distance between 2 parallel lines $Ax+By=C_1=0\&Ax+By+C_2=0$ is given by $\frac{|C_2-C_1|}{\sqrt{A^2+B^2}}$

$\frac{k-a}{5}=\pm \frac{32a}{15}$

$k=a\pm \frac{32a}{3}$

We get 2 values of k, one value for above $l_2$ and other for below, we see that the value required is k for line above $l_1\because l_2$ is in direction of centre relative to $l_1$ and $y_{centre}>y_{F_1}$

or you can check by drawing graph

$\therefore k=-\frac{-29a}{3}$

Equation of directrix is $12x-9y+29a=0$

Centre of hyperbola is at $(-\frac{a}{3},a)$

Equation of hyperbola is $7y^2+15a^2+24xy-24ax-6ay=0$